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Solution: We have solved this problem in the past by assuming
that the object slid without friction along the
track. The condition for maintaining contact with
the track is unchanged from that problem. At every
point along the path, the forces acting on the ball
(the gravitational force, the normal force of the
track, and the frictional force) must act so as to
satisfy the centripetal condition. Since the velocity
of the center-of-mass will be lowest at the top of
the loop and the normal and gravitational forces will
both be pointing down there, this point sets the
condition for maintaining contact with the track
(see the figure below).
To maintain a circular path along the track, at the
top of the loop we must have
The minimum height, h, corresponds to the minimum velocity at the top of the loop since we know it is the work done by gravity that imparts kinetic energy to the ball. In turn, the minimum velocity at the top of the loop corresponds to FN = 0. So, mvtop, min2 = mgR ==> vtop, min2 = gR.
For rolling without slipping, mechanical energy is
conserved (since static friction on the ball does no
work), so, at the top of the loop
| Wgrav | = Ktrans. + Krot. |
| mg(h - 2R) | = ½ mvtop2 + ½ I*omegatop2 |
For the minimum height, h, we need to set
vtop to vtop, min
and to note that v = omega*r for rolling
without slipping. Thus,
| mg(h - 2R) | = ½ mvtop, min2 + ½ I*omegatop2 | = ½ mvtop, min2 + ½ (2/5 mr2)(vtop, min/r)2 |
| mg(h - 2R) | = 7/10 mvtop, min2 |
| g(h - 2R) | = 7/10 gR ==> |
| h | = 7/10 R + 2R = 2.7R |
We note that this result for the minimum h is higher than in the case of a sliding object (2.5 R) since some of the work done by gravity goes into rotating the ball, and the center-of-mass therefore goes more slowly for rolling than for sliding.
Solution: First, we need to make some reasonable assumptions
to simplify the problem. We assume that the force of
the cue stick on the ball is large and over a very
short period of time so that, for this time period,
the change in the translational motion of the ball
is determined exclusively by the cue stick force.
In other words, we will ignore the frictional force
acting on the ball during the time the cue stick
is acting. Second, since rolling without slipping
is to begin immediately, we assume the instantaneous
acceleration of the center-of-mass, a, is equal to
alpha*R. If we consider the torque about an axis
through the center-of-mass and perpendicular to the
page, then, assuming that the frictional force is
negligible for the short time that the cue stick
force is acting, we can get the torque acting about
this axis from the diagram below
Only the component of the force which is perpendicular
to the line connecting the point where the force acts
and the axis of rotation contributes to the angular
acceleration. This component is shown as
Fx. The torque is
| Fsin(theta)*R | = I*alpha |
| F*R*sin(theta) | = (2/5)MR2*a/R |
| Fh | = (2/5)MRa |
The acceleration of the center-of-mass during the
time F is acting is
a = F/M if we assume
the frictional force is negligible during this time.
So, using this in our previous result
| Fh | = = (2/5)MRa |
| = (2/5)MR*(F/M) ==> | |
| h | = 2/5 R |
So we want to hit the cue ball at a point 2/5 R above the center. Hitting below this point causes the ball to slide some distance before rolling without slipping. Hitting above this point causes the cue ball to have top spin. It spins faster than it should and has to slow its rotation rate before rolling without slipping can begin.
Solution: First, we note from our previous problem that if we hit
the ball at a height less than 2/5 R above the center,
then rolling does not begin immediately. In fact, if
we hit the ball dead center, then the ball begins sliding
with no initial angular velocity, hence omega0 = 0.
As the ball slides, the only force acting in the horizontal
direction on it is the friction. This frictional force causes
a torque fR = I*alpha. We can
not assume that alpha = a/R while the ball is sliding,
however, the condition for rolling without slipping to begin
is the time at which v = omega*R. Thus the initial velocity
and angular velocity are v0 and 0, respectively,
and the final velocity and angular velocity are v and omega
where v = omega*R.
The angular velocity as a function of time can be found
once the angular acceleration is known, so
| fR | = I*alpha |
| muMgR | = (2/5)MR2alpha ==> |
| alpha | = (5/2)mug/R |
Let's say that rolling without slipping begins at
a time t. The angular velocity after a time t is
But, by definition of t, omega = v/R, so
Now, the translation equation of motion gives us
the deceleration of the center-of-mass velocity
Since this acceleration is constant, we can use one
of the equations for motion to relate the initial and
final velocities
| v | = v0 + at |
| (5/2)mugt | = v0 - mugt ==> |
| (7/2)mugt | = v0 ==> |
| t | = 2v0/(7mug) |
The distance traveled in this time is
| d | = v0t - ½ at2 |
| = v0[2v0/(7mug)] - ½ (mug/2) [2v0/(7mug)]2 ==> | |
| = 2v02/(7mug) - (2/49) [v02/(7mug] ==> | |
| d | = 12v02/(49mug) |
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Up: Completing the Circle
Previous: Rolling Motion
larryg@upenn5.hep.upenn.edu